Question

A block is shot with an initial velocity of $5\text{m/s}$ on a rough horizontal plane. Find the distance covered by the block till it comes to rest. Given the coefficient of friction between the surfaces of contact is $0.2\text{and}g=10m/s^2$.

• A. $12\text{m}$
• B. $25\text{m}$
• C. $6.25\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is C $6.25\text{m}$

The given situation is shown below


The block will be under the impact of frictional force after getting shot. This will cause retardation. Hence, the acceleration of the block is given by
$F=ma=-f$
Since block is moving, it will experience frictional force
$f=\mu N$

i.e $-\mu N=ma$
$\Rightarrow -\mu mg=ma$
$\Rightarrow a=-\mu g=0.2\times 10=-2{\text{m/s}}^2$

Given, $u=5\text{m/s},a=-2{\text{m/s}}^2,\text{and}v=0$ (since block comes to rest)
From the 3rd equation of motion, we have
$v^2-u^2=2as\Rightarrow 0^2-5^2=-2\left(2\right)\left(s\right)$
$\Rightarrow s=6.25\text{m}$ is the distance covered by the block before stopping.